To do this, we plug in $t=2$ into the equation:
$Q(2) = 1000e^{0.4(2)} = 1000e^{0.8}$
Now, we can use a calculator to find the value of $e^{0.8}$:
$e^{0.8} \approx 2.22554$
So, the number of bacteria present at the end of 2 hours is:
$Q(2) = 1000(2.22554)
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